NCERT Solutions for Exercise 8.4 Class 10 Maths Chapter 8 - Introduction to Trigonometry

NCERT Solutions for Exercise 8.4 Class 10 Maths Chapter 8 - Introduction to Trigonometry

Updated on 30 Apr 2025, 03:22 PM IST

Trigonometric identities provide a foundation that enables quick solutions to multiple mathematical expressions. This exercise demonstrates applying basic trigonometric ratios of sine, cosine and tangent when performing standard identity simplifications and verifications. The linkage among different trigonometric functions becomes possible through these identities, which enable us to show complex expressions in separate steps. Learning trigonometric connections helps students develop their logical reasoning skills, leading to advanced trigonometric abilities needed for engineering, architecture and physical science fields.

This Story also Contains

  1. NCERT Solutions Class 10 Maths Chapter 8: Exercise 8.3
  2. Access Solution of Introduction to Trigonometry Class 10 Chapter 8 Exercise: 8.3
  3. Topics covered in Chapter 8 Introduction to Trigonometry: Exercise 8.3
  4. NCERT Solutions of Class 10 Subject Wise
  5. NCERT Exemplar Solutions of Class 10 Subject-Wise
NCERT Solutions for Exercise 8.4 Class 10 Maths Chapter 8 - Introduction to Trigonometry
exercise 8.4

Exercise 8.3 in the NCERT Solutions for Class 10 presents three important identities which state \(\sin^2 \theta + \cos^2 \theta = 1\) alongside \(1 + \tan^2 \theta = \sec^2 \theta\) and \(1 + \cot^2 \theta = \csc^2 \theta\). The relationships teach students methods to simplify problems through systematic verification of given expressions. The exercise functions as an essential tool for solidifying the knowledge described in NCERT Books. This exercise represents a fundamental requirement for solving trigonometric problems, which progress into height and distance calculations and advanced mathematical concepts.

NCERT Solutions Class 10 Maths Chapter 8: Exercise 8.3

Download PDF


Access Solution of Introduction to Trigonometry Class 10 Chapter 8 Exercise: 8.3

Q 1: Express the trigonometric ratios sinA,secA and tanA in terms of cotA .

Answer:

We know that csc2Acot2A=1
(i)
1sin2A=1+cot2Asin2A=11+cot2AsinA=11+cot2A

(ii) We know the identity of
sec2Atan2A=11cos2A=1+tan2A=1+1cot2Acot2A1+cot2A=cos2AcotA1+cot=cosA

(iii) tanA=1cotA

Q 2: Write all the other trigonometric ratios of A in terms of secA .

Answer:

We know that the identity sin2A+cos2=1

sin2A=1cos2sin2A=11sec2A

=sec2A1sec2A

sinA=sec2A1sec2A

=sec2A1secA

cosecA=secAsec2A1

tanA=sinAcosA=sec2A1

cotA=1sec2A1

Q 3: Choose the correct option. Justify your choice (i)9sec2A9tan2A=

(A) 1 (B) 9 (C) 8 (D) 0

Answer:

The correct option is (B) = 9

9sec2A9tan2A=9(sec2Atan2A) .............(i)

And it is known that sec2Atan2A=1

Therefore, equation (i) becomes, 9×1=9

Q 3: Choose the correct option. Justify your choice (ii)(1+tanθ+secθ)(1+cotθcosecθ)=

(A) 0 (B) 1 (C) 2 (D) –1

Answer:

The correct option is (C)

(1+tanθ+secθ)(1+cotθcosecθ) .......................(i)

we can write his above equation as;

=(1+sinθ/cosθ+1/cosθ)(1+cosθ/sinθ1/sinθ)=(1+sinθ+cosθ)cosθ.sinθ×((sinθ+cosθ1sinθ.cosθ)=(sinθ+cosθ)212sinθ.cosθ=sin2θ+cos2θ+2sinθ.cosθ1sinθ.cosθ=2×sinθ.cosθsinθ.cosθ = 2

Q 3: Choose the correct option. Justify your choice (iii)(secA+tanA)(1sinA)=

(A)secA (B)sinA (C)cosecA (D)cosA

Answer:

The correct option is (D)

(secA+tanA)(1sinA)=

(1cosA+sinAcosA)(1sinA)1+sinAcosA(1sinA)1sin2AcosAcosA

Q 3: Choose the correct option. Justify your choice (iv)1+tan2A1+cot2A=

(A)sec2A (B)1 (C)cot2A (D)tan2A

Answer:

The correct option is (D)

1+tan2A1+cot2A ..........................eq (i)

The above equation can be written as;

We know that cotA=1tanA

therefore,

1+tan2A1+1tan2Atan2A×(1+tan2A1+tan2A)tan2A

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined (i)(cscθcotθ)2=1cosθ1+cosθ

Answer:

We need to prove-

(cscθcotθ)2=1cosθ1+cosθ

Now, taking LHS,

(cscθcotθ)2=(1sinθcosθsinθ)2

=(1cosθsinθ)2=(1cosθ)(1cosθ)sin2θ

=(1cosθ)(1cosθ)1cos2θ=(1cosθ)(1cosθ)(1cosθ)(1+cosθ)=1cosθ1+cosθ

LHS = RHS

Hence proved.

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined (ii)cosA1+sinA+1+sinAcosA=2secA

Answer:

We need to prove-

cosA1+sinA+1+sinAcosA=2secA

Taking LHS;

=cos2A+1+sin2A+2sinAcosA(1+sinA)=2(1+sinA)cosA(1+sinA)=2/cosA=2secA

= RHS

Hence proved.

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined (iii)tanθ1cotθ+cotθ1tanθ=1+secθcscθ

[ Hint: Write the expression in terms of sinθ and cosθ ]

Answer:

We need to prove-

tanθ1cotθ+cotθ1tanθ=1+secθcosecθ

Taking LHS;

tan2θtanθ1+1tanθ(1tanθ) tan3θtan4θ+tanθ1(tanθ1).tanθ.(1tanθ)(tan3θ1)(1tanθ)tanθ.(tanθ1)(1tanθ)

By using the identity a 3 - b 3 =(a - b) (a 2 + b 2 +ab)

(tanθ1)(tan2θ+1+tanθ)tanθ(tanθ1a)tanθ+1+1tanθ1+1+tan2θtanθ1+sec2θ×1tanθ1+secθ.cscθ

= RHS

Hence proved.

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined (iv)1+secAsecA=sin2A1cosA

[ Hint : Simplify LHS and RHS separately]

Answer:

We need to prove-

1+secAsecA=sin2A1cosA

Taking LHS;

1+secAsecA(1+1cosA)/secA1+cosA

Taking RHS;

We know that identity 1cos2θ=sin2θ

sin2A1cosA1cos2A1cosA(1cosA)(1+cosA)(1cosA)1+cosA

LHS = RHS

Hence proved.

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined (v)cosAsinA+1cosA+sinA1=cscA+cotA , using the identity csc2A=1+cot2A

Answer:

We need to prove -

cosAsinA+1cosA+sinA1=cosecA+cotA

Dividing the numerator and denominator by sinA , we get;

=cotA1+cscAcotA+1cscA=(cotA+cscA)(csc2Acot2A)cotA+1cscA=(cscA+cotA)(1cscA+cotA)cotA+1cscA=cscA+cotA=RHS

Hence Proved.

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined (vi)1+sinA1sinA=secA+tanA

Answer:

We need to prove -

1+sinA1sinA=secA+tanA

Taking LHS;

By rationalising the denominator, we get;

=1+sinA1sinA×1+sinA1+sinA=(1+sinA)21sin2A=1+sinAcosA=secA+tanA=RHS

Hence proved.

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (vii)sinθ2sin3θ2cos3θcosθ=tanθ

Answer:

We need to prove -

sinθ2sin3θ2cos3θcosθ=tanθ

Taking LHS;

[we know the identity cos2θ=2cos2θ1=cos2θsin2θ ]

sinθ(12sin2θ)cosθ(2cos2θ1)sinθ(sin2θ+cos2θ2sinθ)cosθ.cos2θsinθ.cos2θcosθ.cos2θtanθ=RHS

Hence proved.

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined (viii)(sinA+cscA)2+(cosA+secA)2=7+tan2A+cot2A.

Answer:

Given the equation,

(sinA+cscA)2+(cosA+secA)2=7+tan2A+cot2A ..................(i)

Taking LHS;

(sinA+cscA)2+(cosA+secA)2

sin2A+csc2A+2+cos2A+sec2A+21+2+2+(1+cot2A)+(1+tan2A)

[since sin2θ+cos2θ=1,csc2θcot2θ=1,sec2θtan2θ=1 ]

7+csc2A+tan2A=RHS

Hence proved

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined (ix)(cosecAsinA)(secAcosA)=1tanA+cotA

[ Hint : Simplify LHS and RHS separately]

Answer:

We need to prove-

(coescAsinA)(secAcosA)=1tanA+cotA

Taking LHS;

(cosecA1cscA)(secA1secA)(cosec21)cosecA×sec2A1secAcot2AcosecA.tan2AsecAsinA.cosA

Taking RHS;

1sinA/cosA+cosA/sinAsinA.cosAsin2A+cos2AsinA.cosA

LHS = RHS

Hence proved.

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined (x)(1+tan2A1+cot2A)=(1tanA1cotA)2=tan2A

Answer:

We need to prove,

(1+tan2A1+cot2A)=(1tanA1cotA)2=tan2A

Taking LHS;

1+tan2A1+cot2A=sec2Acsc2A=tan2A

Taking RHS;

=(1tanA1cotA)2=(1sinA/cosA1cosA/sinA)2=(cosAsinA)2(sin2A)(sinAcosA)2(cos2A)=tan2A

LHS = RHS

Hence proved.


Also read-

Aakash Repeater Courses

Take Aakash iACST and get instant scholarship on coaching programs.

Topics covered in Chapter 8 Introduction to Trigonometry: Exercise 8.3

1. Understanding basic trigonometric identities: You must learn the essential three trigonometric identities and understand their value for problem simplification and solution.

2. Verifying trigonometric identities: Perform proofs which establish two sides of an equation to be equal by using proper transformations in combination with trigonometric relations.

3. Developing logical and proof-solving skills: Step-by-step logical verification belongs to the set of skills that students need to develop for checking identities and producing mathematical proofs.

4. Preparing for advanced applications: Construct a strong foundation to solve problems utilising height-distance relationships alongside trigonometric equation methods, as well as actual angle measurement scenarios.

Check Out-

NCERT Solutions of Class 10 Subject Wise

Students must check the NCERT solutions for class 10 of the Mathematics and Science Subjects.

NCERT Exemplar Solutions of Class 10 Subject-Wise

Students must check the NCERT Exemplar solutions for class 10 of the Mathematics and Science Subjects.

Frequently Asked Questions (FAQs)

Q: Can I use a calculator for solving trigonometric identities in the exam?
A:

No, calculators are not allowed in Class 10 board exams, so you must rely on your memorised values and identities.

Q: Are the questions in Exercise 8.3 important for board exams?
A:

Yes, Exercise 8.3 is very important as identity-based questions frequently appear in CBSE Class 10 board exams

Q: Do I need to memorize all trigonometric identities?
A:

You should at least remember the fundamental identities and standard angle values to solve Exercise 8.3 effectively.

Articles
|
Next
Upcoming School Exams
Ongoing Dates
UP Board 12th Others

10 Aug'25 - 1 Sep'25 (Online)

Ongoing Dates
UP Board 10th Others

11 Aug'25 - 6 Sep'25 (Online)

Certifications By Top Providers
Explore Top Universities Across Globe

Questions related to CBSE Class 10th

On Question asked by student community

Have a question related to CBSE Class 10th ?

Hello,

Yes, you can give the CBSE board exam in 2027.

If your date of birth is 25.05.2013, then in 2027 you will be around 14 years old, which is the right age for Class 10 as per CBSE rules. So, there is no problem.

Hope it helps !

Hello! If you selected “None” while creating your APAAR ID and forgot to mention CBSE as your institution, it may cause issues later when linking your academic records or applying for exams and scholarships that require school details. It’s important that your APAAR ID correctly reflects your institution to avoid verification problems. You should log in to the portal and update your profile to select CBSE as your school. If the system doesn’t allow editing, contact your school’s administration or the APAAR support team immediately so they can correct it for you.

Hello Aspirant,

Here's how you can find it:

  • School ID Card: Your registration number is often printed on your school ID card.

  • Admit Card (Hall Ticket): If you've received your board exam admit card, the registration number will be prominently displayed on it. This is the most reliable place to find it for board exams.

  • School Records/Office: The easiest and most reliable way is to contact your school office or your class teacher. They have access to all your official records and can provide you with your registration number.

  • Previous Mark Sheets/Certificates: If you have any previous official documents from your school or board (like a Class 9 report card that might have a student ID or registration number that carries over), you can check those.

Your school is the best place to get this information.

Hello,

It appears you are asking if you can fill out a form after passing your 10th grade examination in the 2024-2025 academic session.

The answer depends on what form you are referring to. Some forms might be for courses or examinations where passing 10th grade is a prerequisite or an eligibility criteria, such as applying for further education or specific entrance exams. Other forms might be related to other purposes, like applying for a job, which may also have age and educational requirements.

For example, if you are looking to apply for JEE Main 2025 (a competitive exam in India), having passed class 12 or appearing for it in 2025 are mentioned as eligibility criteria.

Let me know if you need imformation about any exam eligibility criteria.

good wishes for your future!!

Hello Aspirant,

"Real papers" for CBSE board exams are the previous year's question papers . You can find these, along with sample papers and their marking schemes , on the official CBSE Academic website (cbseacademic.nic.in).

For notes , refer to NCERT textbooks as they are the primary source for CBSE exams. Many educational websites also provide chapter-wise revision notes and study material that align with the NCERT syllabus. Focus on practicing previous papers and understanding concepts thoroughly.